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Thread: Good with integrals?

  1. #1
    Mar 2008

    Good with integrals?

    A 0.1-F capacitor measures 150V across it. At t=0, the capacitor is connected to a source that sends current i=16t/square root of (4t^2+9) amperes through the circuit. Find the voltage across the capacitor when t=2s.

    I think you use this equation, but im not sure.
    Vc=(1/C)xintegral of ixdt
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  2. #2
    Junior Member
    Apr 2008
    Pittsburgh, PA
    You're completely right.

    Basically, you know the Capacitance, and you're giving the function of current with respect to time. Since $\displaystyle I=\frac{\delta Q}{\delta t}$ you can just Integrate your I function with respect to t.

    $\displaystyle \int_0^2{\frac{16t}{\sqrt{4t^2+9}}}$

    Just use a simple u substitution so that

    $\displaystyle u=4t^2+9$ and $\displaystyle \delta u=8t$

    From there, you can just integrate the function as $\displaystyle 2\int_0^2{\frac{1}{\sqrt{u}}}$

    and you will end up with the answer $\displaystyle 4\sqrt{u}$

    Change your $\displaystyle u$ back into $\displaystyle 4t^2+9$

    Once you plug in your limits of 0 and 2 you will get $\displaystyle \frac{4\sqrt{25}}{C}$

    You know that $\displaystyle C=.1$ so your final result should be $\displaystyle V=200V$
    Last edited by Kalter Tod; Apr 10th 2008 at 01:26 PM.
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